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Math Help - Probability HW due tomorrow.

  1. #1
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    Probability HW due tomorrow.



    I have two;

    1.) A coin is tossed four times, and the sequence of heads to tails is recorded.

    a) List the sequences in the sample space (I am getting 16...is that right?)
    b) A=(at least 3 heads); B=(at most two heads); C=(heads on the third toss); D=(1 head and 3 tails)

    Find: p(A), p(A intersection B), p(A intersection C), p(A union C), p(B), p(C), p(D)

    I just don't understand the intersections & unions. Please explain?

    ----

    2.) If p(A)=0.4, p(B)=0.5, and p(A union B)=0.7, find:

    a)P(A intersection B) [[I got 0.2]]
    b) p(A' union B') [Where A' = A complimentary. Please explain, I am lost!]

    Thanks so much.
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  2. #2
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    Quote Originally Posted by amor_vincit_omnia View Post


    I have two;

    1.) A coin is tossed four times, and the sequence of heads to tails is recorded.

    a) List the sequences in the sample space (I am getting 16...is that right?)

    Mr F says: Yes.

    b) A=(at least 3 heads); B=(at most two heads); C=(heads on the third toss); D=(1 head and 3 tails)

    Find: p(A), p(A intersection B), p(A intersection C), p(A union C), p(B), p(C), p(D)

    I just don't understand the intersections & unions. Please explain?

    Mr F says: Apply the definitions and use the sample space.

    ----

    2.) If p(A)=0.4, p(B)=0.5, and p(A union B)=0.7, find:

    a)P(A intersection B) [[I got 0.2]]

    Mr F says: Correct.

    b) p(A' union B') [Where A' = A complimentary. Please explain, I am lost!]

    Thanks so much.
    For b) I suggest you draw a Karnaugh table to really see what's happening. If I have time I'll start it for you.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    For b) I suggest you draw a Karnaugh table to really see what's happening. If I have time I'll start it for you.
    \, \begin{tabular}{l | c | c | c} & A & A$\, '$ & \\ \hline B & 0.2 & & 0.5 \\ \hline B$\, '$ & & & \\ \hline & 0.4 & & 1 \\ \end{tabular}

    It should be easy to fill in the blanks and then add up the probabilities corresponding to A' or B'.

    Alternatively, Pr(A' U B') = 1 - Pr(A and B) = 1 - 0.2 = 0.8 (of course, this is obvious when looking at the Karnaugh table).
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  4. #4
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    Wow, that one was obvious..LOL. Sorry.

    I am not clear on how to "apply the definitions" to the first problem. Intersection means that both A and B will happen at the same time, or?
    Union means that one or the other happens?
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    Wait, you're saying that p(A' U B') is 0.8?
    Shouldn't it be 1-p(A U B) = 1-.7, or 0.3?
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  6. #6
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    Quote Originally Posted by amor_vincit_omnia View Post
    Wait, you're saying that p(A' U B') is 0.8?
    Shouldn't it be 1-p(A U B) = 1-.7, or 0.3?
    No. Why should it be?

    Fill out the rest of the Karnaugh table and study it carefully. What squares have A' OR B'? Add up the probabilities. Now do you see why Pr(A' U B') = 1 - Pr(A and B) ....?
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  7. #7
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    Quote Originally Posted by amor_vincit_omnia View Post
    [snip]
    Intersection means that both A and B will happen at the same time, or?
    Union means that one or the other happens?
    Yes. So, for example, Pr(A intersection B) = 0 ...... (A and B can't both happen).
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  8. #8
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    I guess I thought that because p(A') = 1- p(A), p(A' U B')= 1-p(A U B)

    I had never seen that chart before, I will look at it.
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  9. #9
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    Quote Originally Posted by amor_vincit_omnia View Post
    I guess I thought that because p(A') = 1- p(A), p(A' U B')= 1-p(A U B)
    [snip]
    If a formula that you invent is not in your textbook or class notes, then Pr(formula is erroneous) --> 1.
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